IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

truly

@ KAB....

u were right by applying N-leibniz rule, but u made an error in integration....see here

u got f '(x) = 1/(f(x))²...this is right.................eqn 1

but whn u integrate LHS f'(x) wrt x is = f(x) ,,,, but whn u integrate 1/(f(x))² wrt x it is NOT -1/f(x) ....

u shud do it like this ...whn u hav got f '(x) = 1/(f(x))²
==> (f(x))² f'(x) = 1 not this can be integrated wrt f'(x)

==> (f(x))³ / 3 = f'(x) + C now putting the value of f'(x) from eqn 1

==> (f(x))³ / 3 = 1/(f(x))² + C

use the condition f(2/3) = 3^{[2 ]}

2 to find C

and then u can calculate f(72)....

This method is complicated .....

again using eqn 1 which was f '(x) = 1/(f(x))²
== > (f(x))² f'(x) = 1 = dM (say)
integrate wrt x

M = x + c1 ......... eqn 2

also M =

(f(x))² f'(x) dx .....use

uv dx

=> M = (f(x))³ -

2f(x)f'(x).f(x) dx = (f(x))³- 2M

=> M = (f(x))³ /3 + c2 ........eqn 3

from eqn 2 & 3

(f(x))³ /3 = x + C

put x = 3/2 ,,,,find C

and thn put x = 72 to find f(72)

I hope i cleared ur doubt where u were wrong

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